poj 2478 Farey Sequence

Description

The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are
F2 = {1/2}
F3 = {1/3, 1/2, 2/3}
F4 = {1/4, 1/3, 1/2, 2/3, 3/4}
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5}

You task is to calculate the number of terms in the Farey sequence Fn.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 10⁶). There are no blank lines between cases. A line with a single 0 terminates the input.

Output

For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn.

Sample Input

2
3
4
5
0

Sample Output

1
3
5
9

题意：给定一个数n，求小于或等于n的数中两两互质组成的真分数的个数。 表达的有点挫，很直接的欧拉函数的应用。

phi(x) 表示与x互质且小于x的正整数的个数。

直接见代码： 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define see(x) cout<<#x<<":"<<x<<endl;
#define MAXN 1000005
using namespace std;

int prime[MAXN],phi[MAXN];
bool notp[MAXN];
void Prime(){
    int i, j, np=0;
    phi[1] = 0;
    memset(notp,false,sizeof(notp));
    for(i=2;i<MAXN;i++){
        if(notp[i]==false){
            prime[np++] = i;
            phi[i] = i-1; 
        }
        for(j=0;j<np&&i*prime[j]<MAXN;j++){
            notp[i*prime[j]] = true;
            if(i%prime[j]==0){
                phi[i*prime[j]] = phi[i]*prime[j];
                break;
            }
            else{
                phi[i*prime[j]] = phi[i]*(prime[j]-1);
            }
        }
    }
}

long long ans[MAXN];
int main(){
    int n, i;
    Prime();
    ans[1] = 0;
    for(i=2;i<MAXN;i++){
        ans[i] = ans[i-1] + phi[i];
    }
    while(scanf("%d",&n)&&n){
        cout<<ans[n]<<endl;
    }
    return 0;
}

这里要说到关于筛素数的方法。以下算法复杂度为O(n)

void Prime(){
    int i, j, np=0;
    memset(notp,false,sizeof(notp));
    for(i=2;i<MAXN;i++){
        if(notp[i]==false){
            prime[np++] = i;
        }
        for(j=0;j<np&&i*prime[j]<MAXN;j++){
            notp[i*prime[j]] = true;
            if(i%prime[j]==0){   //O(n)的关键
                break;
            }
        }
    }
}

适时跳出，大大降低了算法复杂度。本质就是任意一个合数都有一个最小质因数，最终被赋值为true时，都是由其的最小质因数筛出来的，其被筛出来后也无需再关心其他数了。

原因如下：设 i 的最小质因数为p[j]，则 i%p[j]==0，设k=i/p[j]，w=i*p[j]=k*p[j]*p[j]，w是由k*p[j]和p[j]共同筛出来的，对于比w更大的也有因子p[j]的数ww，则自然可以由后面的m*p[j](m>k)和p[j]共同筛出来了。

对筛素数的代码，略加几行代码，就可以顺带求出欧拉函数phi[x]了。用到了递推式：

对于p|x。若p²|x，则phi(x) = phi(x/p)*p；否则phi(x) = phi(x/p)*(p-1)